Context: A quadratic function $g$ models a seal's depth below the surface, in meters, $t$ minutes after entering the water. The seal reached its maximum depth of $409.6$ meters at $t = 8$ and returned to the surface (depth $0$) at $t = 16$. We need the estimated depth at $t = 11$.

Answer: 352

Why: The maximum is the vertex, so write $g$ in vertex form: $g(t) = a(t - 8)^2 + 409.6$. Use $g(16) = 0$: $a(16 - 8)^2 + 409.6 = 0$, so $64a = -409.6$ and $a = -6.4$. Then $g(11) = -6.4(11 - 8)^2 + 409.6 = -6.4(9) + 409.6 = -57.6 + 409.6 = 352$ meters.

Verification: By symmetry about $t = 8$, the parabola should also give depth $0$ at $t = 0$, when the seal entered the water: $g(0) = -6.4(64) + 409.6 = 0$. It does.

Common slips: Forgetting to square $11 - 8 = 3$, which gives $409.6 - 19.2 = 390.4$, or dropping the negative sign on $a$ and reporting a depth greater than the maximum.